The figure shows a particle sliding on a frictionless track, which terminates in a straight horizontal section. If the particle starts slipping from the point A, how far away from the track will the particle hit the ground?

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3 m

4 m

2 m

1 m

answer is A.

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Detailed Solution

The given figure is

From law of conservation of energy, $m g (1) = \frac{1}{2} m v^{2} + m g (0.5)$

$\Rightarrow v = \sqrt{10} m / s$

Time of flight T = $\sqrt{\frac{2 \times 0.5}{10}} = \frac{1}{\sqrt{10}} s$

Therefore R = v $\times T = \frac{1}{\sqrt{10}} \times \sqrt{10} s = 1 m$

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